Powers of Elements in Group Direct Product
Theorem
Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be group whose identities are $e_G$ and $e_H$.
Let $\struct {G \times H, \circ}$ be the group direct product of $G$ and $H$.
Then:
- $\forall n \in \Z: \forall g \in G, h \in H: \tuple {g, h}^n = \tuple {g^n, h^n}$
Proof
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition $\forall g \in G, h \in H: \tuple {g, h}^n = \tuple {g^n, h^n}$.
Basis for the Induction
$\map P 0$ is true, as this says:
- $\tuple {g, h}^0 = \tuple {e_G, e_H}$
$\map P 1$ is true, as this says:
- $\tuple {g, h} = \tuple {g, h}$
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\tuple {g, h}^k = \tuple {g^k, h^k}$
Then we need to show:
- $\tuple {g, h}^{k + 1} = \tuple {g^{k + 1}, h^{k + 1} }$
Induction Step
This is our induction step:
\(\ds \tuple {g, h}^{k + 1}\) | \(=\) | \(\ds \tuple {g, h}^k \circ \tuple {g, h}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {g^k, h^k} \circ \tuple {g, h}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {g^k \circ_1 g, h^k \circ_2 h}\) | Definition of Group Direct Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {g^{k + 1}, h^{k + 1} }\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \N: \forall g \in G, h \in H: \tuple {g, h}^n = \tuple {g^n, h^n}$
So we have shown the result holds true for all $n \ge 0$.
The result for $n < 0$ follows directly from Powers of Group Elements for Negative Indices.
$\blacksquare$